Using Matrix-Chernoff inequality, one can easily show \(\alpha _1^2\ge m\lambda _*/4\)  (see Lemma 5.2 of [1]}) with a probability of at least \(1-\delta \) ,
if \(m= \tilde{\Omega }(N/\lambda _*)\)  holds, where
\(\lambda _*=\lambda _{\min }\left(G_*\right)\)  with \(G_*:=\mathbb {E}_{w\sim \mathcal {N}(0,I_d)}\left[\sigma (Xw)\sigma (Xw)^T\right]\) 
and \(\tilde{\Omega }\)  omits logarithmic factors depending on
\(\delta \) . It can be shown \(\lambda _* > 0\)  under some mild data
assumption. For example, \(\lambda _*>0\)  if no two data are
parallel to each other by Lemma 3.2 of [2]}. Additionally, we have
\(\lambda _*\le \frac{\mathbf {tr}(G_*)}{N}=\frac{\mathbb {E}\Vert \sigma (Xw)\Vert ^2}{N}\le \frac{\mathbb {E}\Vert Xw\Vert ^2}{N}=\frac{\Vert X\Vert _F^2}{N}=1.\) 
